\(\int (g \cos (e+f x))^{-1-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1+n} \, dx\) [186]

Optimal result

Integrand size = 45, antiderivative size = 125 \[ \int (g \cos (e+f x))^{-1-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1+n} \, dx=\frac {(g \cos (e+f x))^{-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1+n}}{f g (2+m-n)}+\frac {(g \cos (e+f x))^{-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n}{c f g (m-n) (2+m-n)} \]

[Out]

(g*cos(f*x+e))^(-n-m)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-1+n)/f/g/(2+m-n)+(g*cos(f*x+e))^(-n-m)*(a+a*sin(f*
x+e))^m*(c-c*sin(f*x+e))^n/c/f/g/(m-n)/(2+m-n)

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.044, Rules used = {2928, 2927} \[ \int (g \cos (e+f x))^{-1-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1+n} \, dx=\frac {(a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{n-1} (g \cos (e+f x))^{-m-n}}{f g (m-n+2)}+\frac {(a \sin (e+f x)+a)^m (c-c \sin (e+f x))^n (g \cos (e+f x))^{-m-n}}{c f g (m-n) (m-n+2)} \]

[In]

Int[(g*Cos[e + f*x])^(-1 - m - n)*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-1 + n),x]

[Out]

((g*Cos[e + f*x])^(-m - n)*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-1 + n))/(f*g*(2 + m - n)) + ((g*Cos[e
+ f*x])^(-m - n)*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^n)/(c*f*g*(m - n)*(2 + m - n))

Rule 2927

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) +
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^
n/(a*f*g*(m - n))), x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] &
& EqQ[m + n + p + 1, 0] && NeQ[m, n]

Rule 2928

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) +
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^
n/(a*f*g*(2*m + p + 1))), x] + Dist[(m + n + p + 1)/(a*(2*m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f
*x])^(m + 1)*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && EqQ[b*c + a*d, 0] &
& EqQ[a^2 - b^2, 0] && ILtQ[Simplify[m + n + p + 1], 0] && NeQ[2*m + p + 1, 0] && (SumSimplerQ[m, 1] ||  !SumS
implerQ[n, 1])

Rubi steps \begin{align*} \text {integral}& = \frac {(g \cos (e+f x))^{-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1+n}}{f g (2+m-n)}+\frac {\int (g \cos (e+f x))^{-1-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx}{c (2+m-n)} \\ & = \frac {(g \cos (e+f x))^{-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1+n}}{f g (2+m-n)}+\frac {(g \cos (e+f x))^{-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n}{c f g (m-n) (2+m-n)} \\ \end{align*}

Mathematica [A] (verified)

Time = 23.90 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.70 \[ \int (g \cos (e+f x))^{-1-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1+n} \, dx=\frac {(g \cos (e+f x))^{-m-n} (a (1+\sin (e+f x)))^m (-1-m+n+\sin (e+f x)) (c-c \sin (e+f x))^n}{c f g (m-n) (2+m-n) (-1+\sin (e+f x))} \]

[In]

Integrate[(g*Cos[e + f*x])^(-1 - m - n)*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-1 + n),x]

[Out]

((g*Cos[e + f*x])^(-m - n)*(a*(1 + Sin[e + f*x]))^m*(-1 - m + n + Sin[e + f*x])*(c - c*Sin[e + f*x])^n)/(c*f*g
*(m - n)*(2 + m - n)*(-1 + Sin[e + f*x]))

Maple [F]

\[\int \left (g \cos \left (f x +e \right )\right )^{-1-m -n} \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{-1+n}d x\]

[In]

int((g*cos(f*x+e))^(-1-m-n)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-1+n),x)

[Out]

int((g*cos(f*x+e))^(-1-m-n)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-1+n),x)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.01 \[ \int (g \cos (e+f x))^{-1-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1+n} \, dx=\frac {{\left ({\left (m - n + 1\right )} \cos \left (f x + e\right ) - \cos \left (f x + e\right ) \sin \left (f x + e\right )\right )} \left (g \cos \left (f x + e\right )\right )^{-m - n - 1} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} e^{\left (2 \, {\left (n - 1\right )} \log \left (g \cos \left (f x + e\right )\right ) - {\left (n - 1\right )} \log \left (a \sin \left (f x + e\right ) + a\right ) + {\left (n - 1\right )} \log \left (\frac {a c}{g^{2}}\right )\right )}}{f m^{2} + f n^{2} + 2 \, f m - 2 \, {\left (f m + f\right )} n} \]

[In]

integrate((g*cos(f*x+e))^(-1-m-n)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-1+n),x, algorithm="fricas")

[Out]

((m - n + 1)*cos(f*x + e) - cos(f*x + e)*sin(f*x + e))*(g*cos(f*x + e))^(-m - n - 1)*(a*sin(f*x + e) + a)^m*e^
(2*(n - 1)*log(g*cos(f*x + e)) - (n - 1)*log(a*sin(f*x + e) + a) + (n - 1)*log(a*c/g^2))/(f*m^2 + f*n^2 + 2*f*
m - 2*(f*m + f)*n)

Sympy [F(-1)]

Timed out. \[ \int (g \cos (e+f x))^{-1-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1+n} \, dx=\text {Timed out} \]

[In]

integrate((g*cos(f*x+e))**(-1-m-n)*(a+a*sin(f*x+e))**m*(c-c*sin(f*x+e))**(-1+n),x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int (g \cos (e+f x))^{-1-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1+n} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate((g*cos(f*x+e))^(-1-m-n)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-1+n),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

Giac [F]

\[ \int (g \cos (e+f x))^{-1-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1+n} \, dx=\int { \left (g \cos \left (f x + e\right )\right )^{-m - n - 1} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (-c \sin \left (f x + e\right ) + c\right )}^{n - 1} \,d x } \]

[In]

integrate((g*cos(f*x+e))^(-1-m-n)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(-1+n),x, algorithm="giac")

[Out]

integrate((g*cos(f*x + e))^(-m - n - 1)*(a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^(n - 1), x)

Mupad [B] (verification not implemented)

Time = 11.28 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.02 \[ \int (g \cos (e+f x))^{-1-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-1+n} \, dx=\frac {{\left (a\,\left (\sin \left (e+f\,x\right )+1\right )\right )}^m\,{\left (-c\,\left (\sin \left (e+f\,x\right )-1\right )\right )}^n\,\left (2\,\cos \left (e+f\,x\right )-\sin \left (2\,e+2\,f\,x\right )+2\,m\,\cos \left (e+f\,x\right )-2\,n\,\cos \left (e+f\,x\right )\right )}{c\,f\,g\,{\left (g\,\cos \left (e+f\,x\right )\right )}^{m+n}\,\left (2\,\cos \left (e+f\,x\right )-\sin \left (2\,e+2\,f\,x\right )\right )\,\left (m^2-2\,m\,n+2\,m+n^2-2\,n\right )} \]

[In]

int(((a + a*sin(e + f*x))^m*(c - c*sin(e + f*x))^(n - 1))/(g*cos(e + f*x))^(m + n + 1),x)

[Out]

((a*(sin(e + f*x) + 1))^m*(-c*(sin(e + f*x) - 1))^n*(2*cos(e + f*x) - sin(2*e + 2*f*x) + 2*m*cos(e + f*x) - 2*
n*cos(e + f*x)))/(c*f*g*(g*cos(e + f*x))^(m + n)*(2*cos(e + f*x) - sin(2*e + 2*f*x))*(2*m - 2*n - 2*m*n + m^2
+ n^2))